Question

find the work done by a force yi plus xj which displace a particle from origin to a point i plus j​

Answer 1

The force acting on the particle is,

$\longrightarrow\vec{\sf{F}}=\sf{y\,\hat i+x\,\hat j}$

The displacement of the particle is,

$\longrightarrow\vec{\sf{r}}=\sf{\hat i+\hat j}$

So the particle is allowed to move from origin (0, 0) to the point (1, 1).

Let the particle move by this force through a small displacement in the plane,

$\longrightarrow\vec{\sf{dr}}=\sf{dx\,\hat i+dy\,\hat j}$

The work done to move the particle through this displacement by the force is,

$\longrightarrow\mathsf{dW}=\vec{\sf{F}}\cdot\vec{\sf{dr}}$

$\longrightarrow\mathsf{dW}=\sf{\left(y\,\hat i+x\,\hat j\right)\cdot\left(dx\,\hat i+dy\,\hat j\right)}$

$\longrightarrow\mathsf{dW}=\sf{y\,dx+x\,dy}$

$\longrightarrow\mathsf{dW}=\sf{d(xy)}$

Hence the total work done to move the particle from (0, 0) to (1, 1) is,

$\displaystyle\longrightarrow\mathsf{W}=\sf{\int\limits_{(0,\ 0)}^{(1,\ 1)}d(xy)}$

$\displaystyle\longrightarrow\mathsf{W}=\sf{\Big[xy\Big]_{(0,\ 0)}^{(1,\ 1)}}$

$\displaystyle\longrightarrow\mathsf{W}=\sf{1\times1-0\times0}$

$\displaystyle\longrightarrow\underline{\underline{\mathsf{W}=\sf{1\ J}}}$

Hence 1 J is the answer.

Answer 2

Answer:

1J

Explanation:

integration dw = integration ( (yi+xj).(dxi+dyj))

where the integrating limit is (0,0) to (1,1)

now it can be easily solved.